Christina DanielThis is an attempt to clarify the brief derivation of the Lorentz transformation in Albert Einstein’s book [1]. Suppose that the speed of light, $ c$, is the same regardless of whether a coordinate system is or is not translating with a nonzero speed.
Next consider two coordinates systems.
The first is called $ K$ and the second is called $ K’$.
The coordinates for $ K$ are $ x,y,z,t$ and the coordinates for $ K’$ are $ x’,y’,z’,t’$.
Let $ K’$ be translating with respect to $ K$ with a speed of $ v$.
The direction of this translation of $ K’$ is along the $ x$-axis of $ K$.
Suppose light flashes at the origin of both coordinate systems at $ t=0=t’$.
The light then travels along the positive $ x$-axis, positive $ x’$-axis, negative $ x$-axis, and the negative $ x’$-axis.
In $ K$ the position of the light is $ X$, and the position of the light in $ K’$ is $ X’$.
Assume that $ t$ and $ t’$ increase from 0.
First consider the situation in $ K$.
At time $ t$, the position of the light along the positive $ x$-axis is $ X = ct.$
Similarly, the position of the light along the negative $ x$-axis is
$ X = -ct.$
These two equations give rise to
$ 0 = X – ct$
and
$ 0 = X + ct.$
A similar analysis can be done for $ K’$.
The key point is that the speed of the light must be the same regardless of whether the coordinate system is translating with a nonzero speed or not.
So it must be true that, if $ X’$ is the distance between the light and the origin of $ K’$,
$ X’ = ct’ $
and
$ X’ = -ct’ $
Equations (4) and (5) give rise to
$ 0 = X’ – ct’. $
and
$ 0 = X’ + ct’. $
Note that equations (3) and (4) can be made more general with variables $ \lambda$ and $ \beta$:
$ 0 = \lambda (X – ct) $
and
$ 0 = \beta(X + ct).$
Since $ 0=0$, perhaps equations (7) and (9) are equal, and perhaps equations (8) and (10) are equal:
$ X’ – ct’ = \lambda (X – ct) $
and
$ X’ + ct’ = \beta(X + ct) $
These equations can be added and subtracted.
Let
$ a \equiv \frac{\lambda + \beta}{2} \in \mathbb{R} $
and
$ b \equiv \frac{\lambda – \beta}{2} \in \mathbb{R}. $
Then
$ X’ = aX – bct $
and
$ ct’ = act – bX.$
Option 1
First consider $ t=0$.
Then equation (15) yields
$ X’ = aX.$
Next consider $ t’=0$ in $ K’$.
Then equation (16) yields
$ \frac{b c}{a} X = c^2 t = vX $
Define a velocity (see Appendix) as
$ v \equiv \frac{bc}{a} \in \mathbb{R}. $
Constrain $ v \ge 0$ to define the physical situation.
Since $ c \ge 0 $, $ b$ and $ a$ must now have the same sign.
Note that equation (17) yields
$ \Delta X’ = a \Delta X $
and if $ \Delta X’ = 1$ then
$ \frac{1}{a} = \Delta X.$
Consider for general $ t$ equation (15):
$ X’ = aX – bct $
along with the previously derived equation with $ t’=0$
$ X = \frac{c^2}{v} t \rightarrow t = \frac{v}{c^2}X $
The previous two equations yield
$ X’ = aX – b \bigg(\frac{v}{c}X \bigg) $
Suppose $ \Delta X = 1$ from the perspective of $ K’$ at $ t’=0$.
Then
$ \Delta X’ = a \bigg( 1 – \frac{v^2}{c^2}\bigg) $
By the principle of relativity the magnitudes of the displacements are equal, so using (21) and equation (25) one has:
$ \bigg| \frac{1}{a} \bigg| = \bigg| a \bigg( 1 – \frac{v^2}{c^2} \bigg) \bigg|.$
Constrain $ v\le c$ so that $ \big( 1 – \frac{v^2}{c^2}\big) \ge 0$. Then
$ a^2 = \frac{1}{1 – \frac{v^2}{c^2}}.$
Therefore,
$ a = \pm \frac{1}{\sqrt{1 – \frac{v^2}{c^2}}}$
and
$ a^2 v^2 = b^2 c^2 \rightarrow b^2 = \frac{a^2 v^2}{c^2}$
Therefore,
$ b = \pm \frac{v}{c} \frac{1}{\sqrt{1 – \frac{v^2}{c^2}}} $
Finally, substitute $ a$ and $ b$ into equations (15) and (16).
$ X’ = \pm \frac{1}{\sqrt{1 – \frac{v^2}{c^2}}}X – \pm \frac{v}{c} \frac{1}{\sqrt{1 – \frac{v^2}{c^2}}}ct $
and
$ ct’ = \pm \frac{1}{\sqrt{1 – \frac{v^2}{c^2}}}ct – \pm \frac{v}{c} \frac{1}{\sqrt{1 – \frac{v^2}{c^2}}}X. $
Choose the positive roots.
$ \boxed{ X’ = \frac{X – vt}{\sqrt{1 – \frac{v^2}{c^2}}}} $
and
$ \boxed{ t’ = \frac{t – \frac{v}{c^2} X }{\sqrt{1 – \frac{v^2}{c^2}}} }.$
Option 2
First consider $ t=0$. Then
$ X’ = aX$
or $ \Delta X’ = a \Delta X.$
Suppose $ \Delta X = 1$.
Then
$ \Delta X’ = a.$
Next consider $ t’=0$.
Then equation (16) yields
$ bX = act.$
Define a velocity $ v \equiv \frac{bc}{a} \in \mathbb{R}$ (see Appendix). Constrain $ v \ge 0$ to define the physical situation.
Since $ c \ge 0 $, $ b$ and $ a$ must now have the same sign.
Then
$ v X = c^2 t $
or
$ t = \frac{vX}{c^2}.$
For general $ t$ equation (15) is
$ X’ = aX – bct $
Substitute the result from $ t’=0$ to obtain
$ X’ = a X \bigg( 1 – \frac{v^2}{c^2} \bigg)$
Hence
$ \Delta X’ = a \Delta X \bigg( 1 – \frac{v^2}{c^2} \bigg)$
Suppose that $ \Delta X’ = 1$.
Then
$ 1 = a \Delta X \bigg( 1 – \frac{v^2}{c^2} \bigg)$
or
$ \frac{1}{a\big( 1 – \frac{v^2}{c^2} \big)} = \Delta X $
From the principle of relativity the magnitudes of the displacements are equal:
$ \bigg| \frac{1}{a\big( 1 – \frac{v^2}{c^2} \big)} \bigg| = |a|.$
Constrain $ v\le c$ so that $ \big( 1 – \frac{v^2}{c^2}\big) \ge 0$. Then
$ \frac{1}{ 1 – \frac{v^2}{c^2} } = a^2.$
Since
$ v^2 = \frac{b^2 c^2}{a^2} \rightarrow \frac{v^2 a^2}{c^2} = b^2 $
one has
$ b = \pm \frac{v }{c} a = \pm \frac{v }{c} \frac{1}{\sqrt{1 – \frac{v^2}{c^2}}}.$
since, from (47), one has
$ a = \pm \frac{1}{\sqrt{1 – \frac{v^2}{c^2}}}.$
Finally, substitute $ a$ and $ b$ back into equations (15) and (16):
$ X’ = \pm \frac{1}{\sqrt{1 – \frac{v^2}{c^2}}}X – \pm \frac{v }{c} \frac{1}{\sqrt{1 – \frac{v^2}{c^2}}}ct $
and
$ ct’ = \pm \frac{1}{\sqrt{1 – \frac{v^2}{c^2}}}ct – \pm \frac{v }{c} \frac{1}{\sqrt{1 – \frac{v^2}{c^2}}} X.$
Choose the positive roots.
$ \boxed{ X’ = \frac{X – vt}{\sqrt{1 – \frac{v^2}{c^2}}} }$
and
$ \boxed{ t’ = \frac{t – \frac{v}{c^2} X}{\sqrt{1 – \frac{v^2}{c^2}}} }.$
Appendix
Recall that
$ X’ = aX – bct.$
Divide by $ a$:
$ \frac{1}{a} X’ – X = – \frac{bc}{a} t $.
Suppose $ X’=0$.
Then, since
$ X = \frac{bc}{a} t = vt $
one has
$ \frac{\Delta X}{\Delta t} = v $
Alternatively, suppose $ X=0$. Then equation (15) yields
$ X’ = -bct $
while equation (16) yields
$ \frac{1}{a}t’ = t .$
Substituting $ t = \frac{t’}{a}$ into $ X’ = -bct$ one obtains
$ X’ = -bc \frac{t’}{a} $
which leads to
$ \Delta X’ = – \frac{bc}{a} \Delta t’ $
or, with $ v \equiv \frac{bc}{a}$,
$ \frac{\Delta X’}{\Delta t’} = -v.$
References
[1] Albert Einstein. Relativity; The Special and General Theory. Appendix A.
Three Rivers Press, 1961.